Let's solve this problem step-by-step.
### Step 1: Calculate the Temperature Change
The initial temperature ([tex]\( T_i \)[/tex]) is [tex]\( 20.5^{\circ}C \)[/tex].
The final temperature ([tex]\( T_f \)[/tex]) is [tex]\( 29.2^{\circ}C \)[/tex].
The temperature change ([tex]\( \Delta T \)[/tex]) is given by:
[tex]\[ \Delta T = T_f - T_i \][/tex]
Substitute the provided temperatures:
[tex]\[ \Delta T = 29.2^{\circ}C - 20.5^{\circ}C \][/tex]
[tex]\[ \Delta T = 8.7^{\circ}C \][/tex]
### Step 2: Calculate the Heat Absorbed by the Water
The mass of water ([tex]\( m_{water} \)[/tex]) is [tex]\( 500 \, \text{g} \)[/tex].
The specific heat capacity of water ([tex]\( C_{H_2O} \)[/tex]) is [tex]\( 4.18 \, J/(g \, ^{\circ}C) \)[/tex].
The heat absorbed by the water ([tex]\( q_{water} \)[/tex]) can be calculated using the formula:
[tex]\[ q_{water} = m_{water} \times C_{H_2O} \times \Delta T \][/tex]
Substitute the values:
[tex]\[ q_{water} = 500 \, \text{g} \times 4.18 \, J/(g \, ^{\circ}C) \times 8.7^{\circ}C \][/tex]
[tex]\[ q_{water} = 18183.0 \, J \][/tex]
### Step 3: Calculate the Heat Absorbed by the Calorimeter
The heat capacity of the dry portion of the calorimeter ([tex]\( C_{calorimeter} \)[/tex]) is [tex]\( 808 \, J / ^{\circ}C \)[/tex].
The heat absorbed by the calorimeter ([tex]\( q_{calorimeter} \)[/tex]) is given by:
[tex]\[ q_{calorimeter} = C_{calorimeter} \times \Delta T \][/tex]
Substitute the values:
[tex]\[ q_{calorimeter} = 808 \, J / ^{\circ}C \times 8.7^{\circ}C \][/tex]
[tex]\[ q_{calorimeter} = 7029.6 \, J \][/tex]
### Step 4: Calculate the Total Heat Absorbed by the System
The total heat absorbed ([tex]\( q_{total} \)[/tex]) is the sum of the heat absorbed by the water and the calorimeter:
[tex]\[ q_{total} = q_{water} + q_{calorimeter} \][/tex]
Substitute the values:
[tex]\[ q_{total} = 18183.0 \, J + 7029.6 \, J \][/tex]
[tex]\[ q_{total} = 25212.6 \, J \][/tex]
### Step 5: Convert the Heat to Kilojoules
[tex]\[ q_{total \, kJ} = \frac{q_{total}}{1000} \][/tex]
Substitute the value:
[tex]\[ q_{total \, kJ} = \frac{25212.6 \, J}{1000} \][/tex]
[tex]\[ q_{total \, kJ} = 25.2126 \, kJ \][/tex]
### Step 6: Calculate the Moles of Propane
The mass of propane ([tex]\( m_{propane} \)[/tex]) is [tex]\( 0.5 \, g \)[/tex].
The molar mass of propane ([tex]\( M_{propane} \)[/tex]) is [tex]\( 44.1 \, g/mol \)[/tex].
The number of moles of propane ([tex]\( n_{propane} \)[/tex]) is given by:
[tex]\[ n_{propane} = \frac{m_{propane}}{M_{propane}} \][/tex]
Substitute the values:
[tex]\[ n_{propane} = \frac{0.5 \, g}{44.1 \, g/mol} \][/tex]
[tex]\[ n_{propane} = 0.011337868480725623 \, mol \][/tex]
### Step 7: Calculate the Enthalpy Change per Mole of Propane
The enthalpy change ([tex]\( \Delta H \)[/tex]) is given by:
[tex]\[ \Delta H = \frac{q_{total \, kJ}}{n_{propane}} \][/tex]
Substitute the values:
[tex]\[ \Delta H = \frac{25.2126 \, kJ}{0.011337868480725623 \, mol} \][/tex]
[tex]\[ \Delta H = 2223.75132 \, kJ/mol \][/tex]
### Final Answer
The enthalpy change of the reaction is [tex]\( -2223.75132 \, kJ/mol \)[/tex]. Note the negative sign, which indicates that the reaction is exothermic (heat is released).
So, the final answer with significant figures is:
[tex]\[ \boxed{-2224 \, kJ/mol} \][/tex]
